This makes sense. Thank you.

If one uses complex numbers it works since i arctan(-iz)=arctanh(z). But my question is why would Mathematica give an answer in complex functions when this is a very simple integral to begin with.

The answer I get is different

What is the answer you got?

I verified it now by hand, Mathematica is correct. I get the answer in terms of $\arctan$, little simpler than what Mathematia gives

$$ I=\int\frac{\sqrt{x^{3}-1}}{x}dx $$

Let $u=x^{3}+1$, then $du=3x^{2}dx$ and the above becomes

$$ I=\int\frac{\sqrt{u}}{3x^{3}}du=\frac{1}{3}\int\frac{\sqrt{u}}{u-1}dx $$

Now let $u=\tan^{2}v$ or $\sqrt{u}=\tan v$, hence $\frac{1}{2}\frac{1} {\sqrt{u}}du=\sec^{2}vdv$ and the above becomes

\begin{align*} I & =\frac{1}{3}\int\frac{\sqrt{u}}{\tan^{2}v-1}\left( 2\sqrt{u}\sec ^{2}v\right) dv\ & =\frac{2}{3}\int\frac{u}{\tan^{2}v-1}\sec^{2}vdv\ & =\frac{2}{3}\int\frac{\tan^{2}v}{\tan^{2}v-1}\sec^{2}vdv \end{align*}

But $\tan^{2}v-1=\sec^{2}v$ hence

\begin{align*} I & =\frac{2}{3}\int\tan^{2}vdv\ & =\frac{2}{3}\left( \tan v-v\right) \end{align*}

Substituting back

$$ I=\frac{2}{3}\left( \sqrt{u}-\arctan\left( \sqrt{u}\right) \right) $$

Substituting back

$$ I=\frac{2}{3}\left( \sqrt{x^{3}+1}-\arctan\left( \sqrt{x^{3}+1}\right)\right) $$